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  • Sequence Number

    题目来源:2017华中农大网络赛G

    Sequence Number

    In Linear algebra, we have learned the definition of inversion number:
    Assuming A is a ordered set with n numbers ( n > 1 ) which are different from eachother. If exist positive integers i , j, ( 1 ≤ i < j ≤ n and A[i] > A[j]), < A[i], A[j]> isregarded as one of A’s inversions. The number of inversions is regarded as inversionnumber. Such as, inversions of array <2,3,8,6,1> are <2,1>, <3,1>, <8,1>, <8,6>,<6,1>,and the inversion number is 5.Similarly, we define a new notion —— equence number, If exist positive integers i, j, ( 1≤ i ≤ j ≤ n and A[i] <= A[j], < A[i], A[j]> is regarded as one of A’s sequence pair. Thenumber of sequence pairs is regarded as sequence number. Define j – i as the length of thesequence pair.Now, we wonder that the largest length S of all sequence pairs for a given array A.

    Input

    There are multiply test cases.
    In each case, the first line is a number N(1<=N<=50000 ), indicates the size of the array,
    the 2th ~n+1th line are one number per line, indicates the element Ai (1<=Ai<=10^9) ofthe array.

    Output

    Output the answer S in one line for each case.

    Sample Input

    5
    2 3 8 6 1

    Sample Output

    3

    题意概括

    给出n个数字,问在这n个数字中存在两个数a[i]<=a[j],求j-i的最大值;

    解题思路

    首先计算出第一个数字差值,查找第二个数字的时候懂i+max开始查找优化,如果不优化的的话,会超时。

    代码

    #include <stdio.h>
    #include <string.h>
    #include <math.h>
    #include <ctype.h>
    #include <queue>
    #include <algorithm>
    
    using namespace std;
    
    
    int a[50010];
    int main ()
    {
        int n,i,j,k,maxx;
        while (scanf("%d",&n)!=EOF)
        {
            k = 0;
            for (i = 1; i <= n; i ++)
                scanf("%d",a+i);
            maxx = 0; j = 1;
            for (i = n; i > 0; i --)
                if (a[i] >= a[1])
                    break;
            maxx = i-1;
            for (j = 2; j <= n-maxx; j ++)
            {
                for (k = n; k >= j+maxx; k --)
                {
                    if (a[j] <= a[k])
                    {
                        maxx = max(maxx,k-j);
                        break;
                    }
                }
            }
            printf("%d
    ",maxx);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/lanaiwanqi/p/10445711.html
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